If $a>0$ and $g\in L^1(-a,a)$, define: $$f:\mathbb{C}\rightarrow\mathbb{C}, z\mapsto \int_{-a}^ag(t)e^{2\pi izt}\operatorname{d}t.$$
Is it true that: $$|f(z)| =o(e^{2\pi a|z|}), z\rightarrow\infty,$$ i.e.: $$\lim_{z\rightarrow\infty} \frac{|f(z)|}{e^{2\pi a|z|}} =0 ?$$
I know that the result is true if $g\in L^p(-a,a)$ with $p>1$, thanks to the answer to this question and the following estimate: $$|f(x+iy)|\le\|g\|_p\left( \int_{-a}^a e^{-2\pi p' yt}\operatorname{d}t \right)^{1/p'} = \|g\|_p \left(\frac{\operatorname{sinh}(2\pi a p' |y|)}{\pi p'|y|} \right)^{1/p'} \le \frac{\|g\|_p}{(\pi p')^{1/p'}} \frac{e^{2\pi a |y|}}{|y|^{1/p'}},$$ so: $$\frac{|f(x+iy)|}{e^{2\pi a |y|}}\le \frac{\|g\|_p}{(\pi p')^{1/p'}} \frac{1}{|y|^{1/p'}}$$ where $p'\in(1,+\infty)$ is such that: $$\frac{1}{p}+\frac{1}{p'}=1.$$
Obviously, if $p=1$, the previous technique fails to give the result I'm looking for, since in this case it gives only that: $$|f(x+iy)|\le \|g\|_1 e^{2\pi a|y|},$$ i.e. an $O-$estimate, so I'm wondering if there is some other technique to get the same result for $p=1$ or if in this case the result is just plainly false.
For any $\epsilon > 0$, there is $0< \delta <a$ such that $\int_{-a}^{-a+\delta} |g| dt + \int_{a-\delta}^a |g| dt < \epsilon$. We have $$f(z) = \int_{-a+ \delta}^{a-\delta} g(t) e^{2\pi i z t} dt + \int_{-a}^{-a+\delta} g(t) e^{2\pi i z t} dt + \int_{a-\delta}^a g(t) e^{2\pi i z t} dt.$$ Hence $$|f(z)| \leq |\int_{-a+ \delta}^{a-\delta} g(t) e^{2\pi i z t} dt| + |\int_{-a}^{-a+\delta} g(t) e^{2\pi i z t} dt| + |\int_{a-\delta}^a g(t) e^{2\pi i z t} dt|.$$ By simple estimates, we have $$|\int_{-a+ \delta}^{a-\delta} g(t) e^{2\pi i z t} dt| \leq e^{2\pi (a-\delta) |z|} \int_{-a}^a |g(t)| dt,$$ $$|\int_{-a}^{-a+\delta} g(t) e^{2\pi i z t} dt| + |\int_{a-\delta}^a g(t) e^{2\pi i z t} dt| \leq e^{2\pi a |z|} \epsilon.$$ Hence $$\limsup_{|z|\to \infty} \frac{|f(z)|}{e^{2\pi a |z|}} \leq \limsup_{|z|\to \infty} (e^{-\delta 2\pi |z|} ||g||_1 + \epsilon) = \epsilon.$$ Since $\epsilon > 0$ is arbitrary, then $$\limsup_{|z|\to \infty} \frac{|f(z)|}{e^{2\pi a |z|}} \leq 0.$$ Evidently, $\liminf_{|z|\to \infty} \frac{|f(z)|}{e^{2\pi a |z|}} \geq 0$. Hence $$\lim_{|z|\to \infty} \frac{|f(z)|}{e^{2\pi a |z|}} = 0.$$ This finishes the proof.